`int(dx)/((x^3-1)^(2/3)x^2)` is equal to

Options

A. `(3)/(4)(1+(1)/(x^(2)))^(4//3)+c`
B. `(x^(3)-1)^(3//2)+c`
C. `3(1+(1)/(x^(3)))^(1//3)+c`
D. `(1-(1)/(x^(3)))^(1//3)+c`

Answer

To solve the integral

\[ I = \int \frac{dx}{(x^3 - 1)^{\frac{2}{3}} x^2} \]

we will follow these steps:

### Step 1: Rewrite the Integral
We can rewrite the integral as:

\[ I = \int \frac{1}{x^2 (x^3 - 1)^{\frac{2}{3}}} \, dx \]

### Step 2: Factor out \(x^3\) from the Denominator
Next, we factor out \(x^3\) from the term \((x^3 - 1)\):

\[ I = \int \frac{1}{x^2 \left(x^3 \left(1 - \frac{1}{x^3}\right)\right)^{\frac{2}{3}}} \, dx \]

This simplifies to:

\[ I = \int \frac{1}{x^2 x^{2} \left(1 - \frac{1}{x^3}\right)^{\frac{2}{3}}} \, dx = \int \frac{1}{x^4 \left(1 - \frac{1}{x^3}\right)^{\frac{2}{3}}} \, dx \]

### Step 3: Substitution
Let \(t = 1 - \frac{1}{x^3}\). Then, we differentiate \(t\):

\[ dt = \frac{3}{x^4} \, dx \quad \Rightarrow \quad dx = \frac{x^4}{3} \, dt \]

### Step 4: Substitute in the Integral
Now substituting \(t\) and \(dx\) into the integral:

\[ I = \int \frac{1}{x^4 t^{\frac{2}{3}}} \cdot \frac{x^4}{3} \, dt = \frac{1}{3} \int \frac{1}{t^{\frac{2}{3}}} \, dt \]

### Step 5: Integrate
Now we integrate:

\[ \int t^{-\frac{2}{3}} \, dt = \frac{t^{\frac{1}{3}}}{\frac{1}{3}} + C = 3t^{\frac{1}{3}} + C \]

Thus,

\[ I = \frac{1}{3} \cdot 3t^{\frac{1}{3}} + C = t^{\frac{1}{3}} + C \]

### Step 6: Substitute Back
Finally, substituting back \(t = 1 - \frac{1}{x^3}\):

\[ I = \left(1 - \frac{1}{x^3}\right)^{\frac{1}{3}} + C \]

### Final Answer
The final result of the integral is:

\[ \int \frac{dx}{(x^3 - 1)^{\frac{2}{3}} x^2} = \left(1 - \frac{1}{x^3}\right)^{\frac{1}{3}} + C \]